| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Calculate Type I error probability |
| Difficulty | Standard +0.8 This is a multi-part hypothesis testing question requiring understanding of Type I and Type II errors, conditional probability, and the law of total probability. Part (iii) is particularly challenging as it requires constructing a probability tree with multiple scenarios and careful tracking of when p changes, going beyond standard textbook exercises. |
| Spec | 2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(4.81\%\) or \(0.0481\) | B1 | One of these only, or more SF. \(N(18,\ 7.2) \to 0.0468\): B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\geq 14) = 0.7077\) | M1, A1 | Allow M1 for answer 0.5722 or 0.8192. 0.708 or 0.7077 or more SF. \(0.2923: 0\); \(N(15,\ 7.5) \to 0.78\): M1A1; \(0.8194\) or \(0.7674\): M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Only way that \(p = 0.5\) for second test is if Type II error on first, where \(0.2 \times 0.7077 = 0.14154\). Therefore \(0.14154 \times 0.2923 + 0.85846 \times 0.0481 = \mathbf{0.0827}\) | M1, M1, M2, A1 | \(0.2 \times 0.7077 \times 0.2923\ [= 0.04137]\). Consider \(1 - 0.14154\). \(0.2\times(\text{ii})\times(1-(\text{ii})) + (1-[0.2\times(\text{ii})])\times(\text{i})\ [= 0.04137 + 0.04127]\). Answer a.r.t. 0.083 |
| OR: \(0.8\times0.0481\times\overline{0.0481}\ [0.00185] + 0.8\times0.9519\times0.0481\ [0.03663] + 0.2\times0.2923\times0.0481\ [0.00281] + 0.2\times\overline{0.7077}\times0.2923\ [0.04137]\); Add up 4 terms of 3 multiplications; Answer 0.0827 | M1, M1, M1, M1, A1 | Any two of these three: M1; third of these three: M1; this one: M1. SR: No 0.8 or 0.2 but 2 products: M1; 4 products: M2 |
## Question 9:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4.81\%$ or $0.0481$ | B1 | One of these only, or more SF. $N(18,\ 7.2) \to 0.0468$: B1 |
**[1 mark]**
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\geq 14) = 0.7077$ | M1, A1 | Allow M1 for answer 0.5722 or 0.8192. 0.708 or 0.7077 or more SF. $0.2923: 0$; $N(15,\ 7.5) \to 0.78$: M1A1; $0.8194$ or $0.7674$: M1A0 |
**[2 marks]**
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Only way that $p = 0.5$ for second test is if Type II error on first, where $0.2 \times 0.7077 = 0.14154$. Therefore $0.14154 \times 0.2923 + 0.85846 \times 0.0481 = \mathbf{0.0827}$ | M1, M1, M2, A1 | $0.2 \times 0.7077 \times 0.2923\ [= 0.04137]$. Consider $1 - 0.14154$. $0.2\times(\text{ii})\times(1-(\text{ii})) + (1-[0.2\times(\text{ii})])\times(\text{i})\ [= 0.04137 + 0.04127]$. Answer a.r.t. 0.083 |
| OR: $0.8\times0.0481\times\overline{0.0481}\ [0.00185] + 0.8\times0.9519\times0.0481\ [0.03663] + 0.2\times0.2923\times0.0481\ [0.00281] + 0.2\times\overline{0.7077}\times0.2923\ [0.04137]$; Add up 4 terms of 3 multiplications; Answer 0.0827 | M1, M1, M1, M1, A1 | Any two of these three: M1; third of these three: M1; this one: M1. SR: No 0.8 or 0.2 but 2 products: M1; 4 products: M2 |
**[5 marks]**
# Appendix 2 – Question 6(i) Specific Examples9 The random variable $A$ has the distribution $\mathrm { B } ( 30 , p )$. A test is carried out of the hypotheses $\mathrm { H } _ { 0 } : p = 0.6$ against $\mathrm { H } _ { 1 } : p < 0.6$. The critical region is $A \leqslant 13$.\\
(i) State the probability that $\mathrm { H } _ { 0 }$ is rejected when $p = 0.6$.\\
(ii) Find the probability that a Type II error occurs when $p = 0.5$.\\
(iii) It is known that on average $p = 0.5$ on one day in five, and on other days the value of $p$ is 0.6 . On each day two tests are carried out. If the result of the first test is that $\mathrm { H } _ { 0 }$ is rejected, the value of $p$ is adjusted if necessary, to ensure that $p = 0.6$ for the rest of the day. Otherwise the value of $p$ remains the same as for the first test. Calculate the probability that the result of the second test is to reject $\mathrm { H } _ { 0 }$.
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\hfill \mbox{\textit{OCR S2 2013 Q9 [8]}}